Problem: $f(x) = \dfrac{ 1 }{ \sqrt{ x - 9 } }$ What is the domain of the real-valued function $f(x)$ ?
Solution: First, we need to consider that $f(x)$ is undefined when the radicand (the expression under the radical) is less than zero. So the radicand, $x - 9$ , must be greater than or equal to zero. So $x - 9 \geq 0$ ; this means $x \geq 9$ Next, we also need to consider that $f(x)$ is undefined when the denominator, $\sqrt{ x - 9 }$ , is zero. So $\sqrt{ x - 9 } \neq 0$ $\sqrt{ z } = 0$ only when $z = 0$ , so $\sqrt{ x - 9 } \neq 0$ means that $x - 9 \neq 0$ So $x \neq 9$ So we have two restrictions: $x \geq 9$ and $x \neq 9$ Combining these two restrictions, we are left with simply $x > 9$ Expressing this in mathematical notation, the domain is $\{ \, x \in \RR \mid x >9\, \}$.